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How to prove this relationship for Pearson´s c.c.

$$\frac{E((X-\mu)(Y-\nu))}{\sqrt{E((X-\mu)^{2})E((Y-\nu)^{2})}}=\frac{E((X_{1}-X_{2})(Y_{1}-Y_{2}))}{\sqrt{E((X_{1}-X_{2})^{2})E((Y_{1}-Y_{2})^{2})}},$$

where $(X_{1},Y_{1}),(X_{2},Y_{2})$ are i.i.d. as $(X,Y)$ and $E(X)=\mu, E(Y)=\nu.$

Thanks in advance for any comments.

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Try adding and subtracting the means and then expand the squares and the products, maybe work each expectation apart to see the result. For instance, $$E(X_1-X_2)^2=E\big((X_1-\mu)-(X_2-\mu)\big)^2=$$ $$=E(X_1-\mu)^2-2E(X_1-\mu)(X_2-\mu)+E(X_2-\mu)^2=\cdots$$ and so on. Try to check that in this case you get $2 V(X)$.

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    $\begingroup$ Thank you so much, it rally works. Just need to expand denominator/numerator and use independence of components of these vectors. $\endgroup$ – stanly Nov 30 '17 at 13:25

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